Question Number 157739 by gsk2684 last updated on 27/Oct/21 $${two}\:{parallel}\:{paths}\:\mathrm{26}{m}\:{apart}\:{run} \\ $$$${east}−{west}\:{through}\:{the}\:{woods}. \\ $$$${one}\:{person}\:{east}\:{on}\:{one}\:{path}\:{at}\: \\ $$$$\mathrm{3}\:{kmph}\:{and}\:{another}\:{west}\:{on}\:{other}\: \\ $$$${path}\:{at}\:\mathrm{4}\:{kmph}.\:{if}\:{they}\:{pass}\:{each}\: \\ $$$${other}\:{at}\:{time}\:{t}\:=\mathrm{0}\:,\: \\ $$$$\left.{i}\right){how}\:{far}\:{apart}\:{are}\:{they} \\ $$$$\mathrm{8}\:{sec}\:{later}?\: \\…
Question Number 26652 by ajfour last updated on 27/Dec/17 Commented by ajfour last updated on 27/Dec/17 $${If}\:\:{length}\:\boldsymbol{{c}}\:{is}\:{given}\:{and}\:{it}\:{is} \\ $$$${desired}\:{to}\:{start}\:{from}\:{A}\:{and} \\ $$$${reach}\:{B}\:{along}\:{APB}\:{then}\:{at} \\ $$$${what}\:{angle}\:\boldsymbol{\theta}\:{must}\:{we}\:{proceed}; \\ $$$$\left({construction}\:{method}\:{i}\:{mean}\right).…
Question Number 26644 by tawa tawa last updated on 27/Dec/17 Answered by Rasheed.Sindhi last updated on 27/Dec/17 $$\mathrm{Let}\:\mathrm{the}\:\mathrm{side}\:\mathrm{of}\:\mathrm{square}=\mathrm{x} \\ $$$$\mathrm{The}\:\mathrm{diagonal}=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }=\mathrm{28}\sqrt{\mathrm{2}} \\ $$$$\sqrt{\mathrm{2}}\:\:\mathrm{x}=\mathrm{28}\sqrt{\mathrm{2}} \\…
Question Number 157652 by ajfour last updated on 26/Oct/21 Commented by ajfour last updated on 26/Oct/21 $${just}\:{tried}\:{a}\:{symmetrical}\:{eq}.\:{plot}! \\ $$ Commented by mr W last updated…
Question Number 26389 by ajfour last updated on 24/Dec/17 Commented by ajfour last updated on 24/Dec/17 $$\:\:\:{r}+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }={c}\:;\:\:{or}\:\:\:\mathrm{cos}\:\theta+\mathrm{sec}\:^{\mathrm{2}} \theta\:={c} \\ $$$$\Rightarrow\:\:{if}\:\:{x}=\frac{\mathrm{1}}{{r}}\:\:\:{then}\:\:{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}}={c}\: \\ $$$${or}\:\:\:{x}^{\mathrm{3}} −{cx}+\mathrm{1}=\mathrm{0}…
Question Number 157292 by mr W last updated on 21/Oct/21 Commented by mr W last updated on 21/Oct/21 Commented by mr W last updated on…
Question Number 157225 by mr W last updated on 21/Oct/21 Commented by mr W last updated on 21/Oct/21 $${Solution}\:{to}\:{Q}\mathrm{157114} \\ $$$$ \\ $$$${AD}=\sqrt{\mathrm{15}^{\mathrm{2}} −\left(\mathrm{3}{x}\right)^{\mathrm{2}} }…
Question Number 157114 by amin96 last updated on 20/Oct/21 Commented by mathlove last updated on 20/Oct/21 $${wdzr}\:{from}\:{yoj} \\ $$ Commented by mathlove last updated on…
Question Number 157054 by andradamaryjane last updated on 19/Oct/21 $${suppose}\:{you}\:{drop}\:{a}\:{tennis}\:{ball}\:{from}\:{a}\:{hieght}\:{of}\:\mathrm{15}\:{feet}.{after}\:{the}\:{ballhits}\:{the}\:{floor}\:{it}\:{rebounds}\:\:{to}\mathrm{85\%}\:{of}\:{its}\:{previous}\:{height}.{how}\:{high}\:{will}\:{the}\:{ball}\:{rebound}\:{after}\:{its}\:{ghird}\:{bounce}\:{round}\:{tl}\:{the}\:{nearest}\:{tenth} \\ $$$$ \\ $$ Commented by mr W last updated on 19/Oct/21 $${suppose}\:{you}\:{don}'{t}\:{write}\:{all}\:{text}\:{in}\:{a} \\ $$$${single}\:{line},\:{maybe}\:{more}\:{people}\:{may}…
Question Number 25963 by tawa tawa last updated on 16/Dec/17 $$\mathrm{A}\:\mathrm{bus}\:\mathrm{is}\:\mathrm{traveling}\:\mathrm{along}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{road}\:\mathrm{at}\:\mathrm{100}\:\mathrm{km}/\mathrm{hr}\:\mathrm{and}\:\mathrm{the}\:\mathrm{bus}\:\mathrm{conductor} \\ $$$$\mathrm{walks}\:\mathrm{at}\:\mathrm{6}\:\mathrm{km}/\mathrm{hr}\:\mathrm{on}\:\mathrm{the}\:\mathrm{floor}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bus}\:\mathrm{and}\:\mathrm{in}\:\mathrm{the}\:\mathrm{same}\:\mathrm{direction}\:\mathrm{as}\:\mathrm{the}\:\mathrm{bus}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{conductor}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{the}\:\mathrm{road}\:\mathrm{and}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{the}\:\mathrm{bus}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com